Maximal compact subgroups
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From: Chip Eastham on 11 Aug 2010 09:51 On Aug 11, 8:59 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > On 11 Ago, 08:13, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > > > > > > > On Tue, 10 Aug 2010 18:19:25 -0700 (PDT), Louis Burnside > > > <burnside.lo...(a)gmail.com> wrote: > > >On 10 Ago, 21:47, Rupert <rupertmccal...(a)yahoo.com> wrote: > > >> On Aug 11, 10:22 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > >> > Hello everybody! > > > >> > A maximal compact subgroup does not exist even for a general Lie > > >> > group. > > >> > However, there is the famous Malcev-Iwasawa theorem > > >> > (e.g. > > > >> >http://books.google.com/books?id=3_BPupMDRr8C&pg=PA263&dq=malcev-iwas... > > > >> > ). > > >> > I want to understand, for instance, the simple case when G is compact > > >> > and totally disconnected: > > >> > Is the existence of a maximal compact subgroup obvious in this case? > > >> > Is there a simple proof? > > > >> > Best wishes, > > >> > Louis. > > > >> Do you perhaps mean "locally compact"? > > > >Hi Rupert! > > > >I really mean compact. > > > Fine. If G is compact then G is a maximal compact subgroup of G. > > > >I just learned that a compact and totally > > >disconnected group is also called a profinite group. > > >So maybe we can somehow build the maximal compact subgroup from the > > >maximal subgroups of the finite quotients of it. > > >I don't know... > > > >Best, > > >Louis. > > What I really meant is maximal compact proper subgroup of G - sorry. > Equivalently, maximal closed proper subgroup if G is compact (and > Hausdorff). > > Best, > Louis. As you have pointed out (if I understood correctly), no thing necessarily exists. Consider for example the unit circle, considered as an abelian group of rotations. Whatever (topologically) closed proper subgroup we consider, there's always a bigger one. On the other hand we can as well cook up an example where a maximal closed proper subgroup of a compact topological group does exist, e.g. the direct sum of the circle S^1 and Z/2Z, in which of course the circle sits as a subgroup of index 2. Perhaps you are looking for examples (probably not the ones I gave, since things are never that simple), or perhaps you are looking for some conditions that guarantee (or exclude) the existence of the sought maximal compact subgroup. I look forward to further clarification of your aims. regards, chip
From: Louis Burnside on 11 Aug 2010 14:50 On 11 Ago, 10:51, Chip Eastham <hardm...(a)gmail.com> wrote: > On Aug 11, 8:59 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > > > > On 11 Ago, 08:13, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > > > > On Tue, 10 Aug 2010 18:19:25 -0700 (PDT), Louis Burnside > > > > <burnside.lo...(a)gmail.com> wrote: > > > >On 10 Ago, 21:47, Rupert <rupertmccal...(a)yahoo.com> wrote: > > > >> On Aug 11, 10:22 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > >> > Hello everybody! > > > > >> > A maximal compact subgroup does not exist even for a general Lie > > > >> > group. > > > >> > However, there is the famous Malcev-Iwasawa theorem > > > >> > (e.g. > > > > >> >http://books.google.com/books?id=3_BPupMDRr8C&pg=PA263&dq=malcev-iwas... > > > > >> > ). > > > >> > I want to understand, for instance, the simple case when G is compact > > > >> > and totally disconnected: > > > >> > Is the existence of a maximal compact subgroup obvious in this case? > > > >> > Is there a simple proof? > > > > >> > Best wishes, > > > >> > Louis. > > > > >> Do you perhaps mean "locally compact"? > > > > >Hi Rupert! > > > > >I really mean compact. > > > > Fine. If G is compact then G is a maximal compact subgroup of G. > > > > >I just learned that a compact and totally > > > >disconnected group is also called a profinite group. > > > >So maybe we can somehow build the maximal compact subgroup from the > > > >maximal subgroups of the finite quotients of it. > > > >I don't know... > > > > >Best, > > > >Louis. > > > What I really meant is maximal compact proper subgroup of G - sorry. > > Equivalently, maximal closed proper subgroup if G is compact (and > > Hausdorff). > > > Best, > > Louis. > > As you have pointed out (if I understood correctly), > no thing necessarily exists. Consider for example > the unit circle, considered as an abelian group of > rotations. Whatever (topologically) closed proper > subgroup we consider, there's always a bigger one. > > On the other hand we can as well cook up an example > where a maximal closed proper subgroup of a compact > topological group does exist, e.g. the direct sum > of the circle S^1 and Z/2Z, in which of course the > circle sits as a subgroup of index 2. > > Perhaps you are looking for examples (probably not > the ones I gave, since things are never that simple), > or perhaps you are looking for some conditions that > guarantee (or exclude) the existence of the sought > maximal compact subgroup. I look forward to further > clarification of your aims. > > regards, chip Thank you all for replying and for the examples! Please let me clearly state the question: "Let G be a compact and totally disconnected group. How can we prove that G has a maximal compact proper subgroup?" An example: if Z_p is the additive group of the p-adic integers, then pZ_p is its maximal compact proper subgroup. Best, Louis.
From: Rupert on 11 Aug 2010 19:53 On Aug 12, 4:50 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > On 11 Ago, 10:51, Chip Eastham <hardm...(a)gmail.com> wrote: > > > > > On Aug 11, 8:59 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > On 11 Ago, 08:13, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > > > > > On Tue, 10 Aug 2010 18:19:25 -0700 (PDT), Louis Burnside > > > > > <burnside.lo...(a)gmail.com> wrote: > > > > >On 10 Ago, 21:47, Rupert <rupertmccal...(a)yahoo.com> wrote: > > > > >> On Aug 11, 10:22 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > > >> > Hello everybody! > > > > > >> > A maximal compact subgroup does not exist even for a general Lie > > > > >> > group. > > > > >> > However, there is the famous Malcev-Iwasawa theorem > > > > >> > (e.g. > > > > > >> >http://books.google.com/books?id=3_BPupMDRr8C&pg=PA263&dq=malcev-iwas... > > > > > >> > ). > > > > >> > I want to understand, for instance, the simple case when G is compact > > > > >> > and totally disconnected: > > > > >> > Is the existence of a maximal compact subgroup obvious in this case? > > > > >> > Is there a simple proof? > > > > > >> > Best wishes, > > > > >> > Louis. > > > > > >> Do you perhaps mean "locally compact"? > > > > > >Hi Rupert! > > > > > >I really mean compact. > > > > > Fine. If G is compact then G is a maximal compact subgroup of G. > > > > > >I just learned that a compact and totally > > > > >disconnected group is also called a profinite group. > > > > >So maybe we can somehow build the maximal compact subgroup from the > > > > >maximal subgroups of the finite quotients of it. > > > > >I don't know... > > > > > >Best, > > > > >Louis. > > > > What I really meant is maximal compact proper subgroup of G - sorry. > > > Equivalently, maximal closed proper subgroup if G is compact (and > > > Hausdorff). > > > > Best, > > > Louis. > > > As you have pointed out (if I understood correctly), > > no thing necessarily exists. Consider for example > > the unit circle, considered as an abelian group of > > rotations. Whatever (topologically) closed proper > > subgroup we consider, there's always a bigger one. > > > On the other hand we can as well cook up an example > > where a maximal closed proper subgroup of a compact > > topological group does exist, e.g. the direct sum > > of the circle S^1 and Z/2Z, in which of course the > > circle sits as a subgroup of index 2. > > > Perhaps you are looking for examples (probably not > > the ones I gave, since things are never that simple), > > or perhaps you are looking for some conditions that > > guarantee (or exclude) the existence of the sought > > maximal compact subgroup. I look forward to further > > clarification of your aims. > > > regards, chip > > Thank you all for replying and for the examples! > Please let me clearly state the question: > "Let G be a compact and totally disconnected group. > How can we prove that G has a maximal compact proper subgroup?" > > An example: if Z_p is the additive group of the p-adic integers, > then pZ_p is its maximal compact proper subgroup. > > Best, > Louis. Let G be a nontrivial profinite group. Let G_0 be a proper closed subgroup of G and let H be a subgroup of G/G_0 of prime index in G/ G_0. HG_0 is a maximal proper closed subgroup of G, but it is not necessarily unique.
From: Louis Burnside on 12 Aug 2010 21:27 On 11 Ago, 20:53, Rupert <rupertmccal...(a)yahoo.com> wrote: > On Aug 12, 4:50 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > > > > On 11 Ago, 10:51, Chip Eastham <hardm...(a)gmail.com> wrote: > > > > On Aug 11, 8:59 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > > On 11 Ago, 08:13, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > > > > > > On Tue, 10 Aug 2010 18:19:25 -0700 (PDT), Louis Burnside > > > > > > <burnside.lo...(a)gmail.com> wrote: > > > > > >On 10 Ago, 21:47, Rupert <rupertmccal...(a)yahoo.com> wrote: > > > > > >> On Aug 11, 10:22 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > > > >> > Hello everybody! > > > > > > >> > A maximal compact subgroup does not exist even for a general Lie > > > > > >> > group. > > > > > >> > However, there is the famous Malcev-Iwasawa theorem > > > > > >> > (e.g. > > > > > > >> >http://books.google.com/books?id=3_BPupMDRr8C&pg=PA263&dq=malcev-iwas... > > > > > > >> > ). > > > > > >> > I want to understand, for instance, the simple case when G is compact > > > > > >> > and totally disconnected: > > > > > >> > Is the existence of a maximal compact subgroup obvious in this case? > > > > > >> > Is there a simple proof? > > > > > > >> > Best wishes, > > > > > >> > Louis. > > > > > > >> Do you perhaps mean "locally compact"? > > > > > > >Hi Rupert! > > > > > > >I really mean compact. > > > > > > Fine. If G is compact then G is a maximal compact subgroup of G. > > > > > > >I just learned that a compact and totally > > > > > >disconnected group is also called a profinite group. > > > > > >So maybe we can somehow build the maximal compact subgroup from the > > > > > >maximal subgroups of the finite quotients of it. > > > > > >I don't know... > > > > > > >Best, > > > > > >Louis. > > > > > What I really meant is maximal compact proper subgroup of G - sorry.. > > > > Equivalently, maximal closed proper subgroup if G is compact (and > > > > Hausdorff). > > > > > Best, > > > > Louis. > > > > As you have pointed out (if I understood correctly), > > > no thing necessarily exists. Consider for example > > > the unit circle, considered as an abelian group of > > > rotations. Whatever (topologically) closed proper > > > subgroup we consider, there's always a bigger one. > > > > On the other hand we can as well cook up an example > > > where a maximal closed proper subgroup of a compact > > > topological group does exist, e.g. the direct sum > > > of the circle S^1 and Z/2Z, in which of course the > > > circle sits as a subgroup of index 2. > > > > Perhaps you are looking for examples (probably not > > > the ones I gave, since things are never that simple), > > > or perhaps you are looking for some conditions that > > > guarantee (or exclude) the existence of the sought > > > maximal compact subgroup. I look forward to further > > > clarification of your aims. > > > > regards, chip > > > Thank you all for replying and for the examples! > > Please let me clearly state the question: > > "Let G be a compact and totally disconnected group. > > How can we prove that G has a maximal compact proper subgroup?" > > > An example: if Z_p is the additive group of the p-adic integers, > > then pZ_p is its maximal compact proper subgroup. > > > Best, > > Louis. > > Let G be a nontrivial profinite group. Let G_0 be a proper closed > subgroup of G and let H be a subgroup of G/G_0 of prime index in G/ > G_0. HG_0 is a maximal proper closed subgroup of G, but it is not > necessarily unique. Dear Rupert, Every subgroup of prime index surely is maximal. But I don't understand why H exists... I tried a different and very standard approach: Let G be a non-trivial Hausdorff group. Let F be the family of all closed proper subgroups. Order F partially by inclusion. Since {1} belongs to F, F is non-empty. To apply Zorn's Lemma it is enough show that every chain in F has an upper bound in F. Let (H_a lpha) be a chain in F. Let H be the union of all H_alpha, then H is a subgroup (but not necessary closed). Let C be the closure of H in G , then C is a closed subgroup of G. Is C properly contained in G? If so, we are done. Best, Louis.
From: Rupert on 13 Aug 2010 04:21
On Aug 13, 11:27 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > On 11 Ago, 20:53, Rupert <rupertmccal...(a)yahoo.com> wrote: > > > > > On Aug 12, 4:50 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > On 11 Ago, 10:51, Chip Eastham <hardm...(a)gmail.com> wrote: > > > > > On Aug 11, 8:59 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > > > On 11 Ago, 08:13, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > > > > > > > On Tue, 10 Aug 2010 18:19:25 -0700 (PDT), Louis Burnside > > > > > > > <burnside.lo...(a)gmail.com> wrote: > > > > > > >On 10 Ago, 21:47, Rupert <rupertmccal...(a)yahoo.com> wrote: > > > > > > >> On Aug 11, 10:22 am, Louis Burnside <burnside.lo...(a)gmail.com> wrote: > > > > > > > >> > Hello everybody! > > > > > > > >> > A maximal compact subgroup does not exist even for a general Lie > > > > > > >> > group. > > > > > > >> > However, there is the famous Malcev-Iwasawa theorem > > > > > > >> > (e.g. > > > > > > > >> >http://books.google.com/books?id=3_BPupMDRr8C&pg=PA263&dq=malcev-iwas... > > > > > > > >> > ). > > > > > > >> > I want to understand, for instance, the simple case when G is compact > > > > > > >> > and totally disconnected: > > > > > > >> > Is the existence of a maximal compact subgroup obvious in this case? > > > > > > >> > Is there a simple proof? > > > > > > > >> > Best wishes, > > > > > > >> > Louis. > > > > > > > >> Do you perhaps mean "locally compact"? > > > > > > > >Hi Rupert! > > > > > > > >I really mean compact. > > > > > > > Fine. If G is compact then G is a maximal compact subgroup of G.. > > > > > > > >I just learned that a compact and totally > > > > > > >disconnected group is also called a profinite group. > > > > > > >So maybe we can somehow build the maximal compact subgroup from the > > > > > > >maximal subgroups of the finite quotients of it. > > > > > > >I don't know... > > > > > > > >Best, > > > > > > >Louis. > > > > > > What I really meant is maximal compact proper subgroup of G - sorry. > > > > > Equivalently, maximal closed proper subgroup if G is compact (and > > > > > Hausdorff). > > > > > > Best, > > > > > Louis. > > > > > As you have pointed out (if I understood correctly), > > > > no thing necessarily exists. Consider for example > > > > the unit circle, considered as an abelian group of > > > > rotations. Whatever (topologically) closed proper > > > > subgroup we consider, there's always a bigger one. > > > > > On the other hand we can as well cook up an example > > > > where a maximal closed proper subgroup of a compact > > > > topological group does exist, e.g. the direct sum > > > > of the circle S^1 and Z/2Z, in which of course the > > > > circle sits as a subgroup of index 2. > > > > > Perhaps you are looking for examples (probably not > > > > the ones I gave, since things are never that simple), > > > > or perhaps you are looking for some conditions that > > > > guarantee (or exclude) the existence of the sought > > > > maximal compact subgroup. I look forward to further > > > > clarification of your aims. > > > > > regards, chip > > > > Thank you all for replying and for the examples! > > > Please let me clearly state the question: > > > "Let G be a compact and totally disconnected group. > > > How can we prove that G has a maximal compact proper subgroup?" > > > > An example: if Z_p is the additive group of the p-adic integers, > > > then pZ_p is its maximal compact proper subgroup. > > > > Best, > > > Louis. > > > Let G be a nontrivial profinite group. Let G_0 be a proper closed > > subgroup of G and let H be a subgroup of G/G_0 of prime index in G/ > > G_0. HG_0 is a maximal proper closed subgroup of G, but it is not > > necessarily unique. > > Dear Rupert, > > Every subgroup of prime index surely is maximal. But I don't > understand why H exists... > Yes, you are right, this is not clear. I thought it followed from finite group theory. It would be the case when G/G_0 is a p-group. > I tried a different and very standard approach: > Let G be a non-trivial Hausdorff group. > Let F be the family of all closed proper subgroups. Order F partially > by inclusion. Since {1} belongs to F, F is non-empty. > To apply Zorn's Lemma it is enough show that every chain in F has an > upper bound in F. Let (H_a lpha) be a chain in F. > Let H be the union of all H_alpha, then H is a subgroup (but not > necessary closed). > Let C be the closure of H in G , then C is a closed subgroup of G. > Is C properly contained in G? If so, we are done. > This would not be true in general. For example, if G is the absolute Galois group of Q, then G has closed proper subgroups of index 2 and 3 respectively, and their union is all of G. > Best, > Louis. |